The value of the integral $$\int_0^{\infty} e^{-x^2} \, dx$$ is​

Correct option is C

Solution:

​$$\text{We want to evaluate:} \quad \int_0^\infty e^{-x^2} \, dx\\\ \\\text{Let } I = \int_{-\infty}^{\infty} e^{-x^2} \, dx\\\ \\\text{Then } I^2 = \left( \int_{-\infty}^{\infty} e^{-x^2} \, dx \right)^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy\\\ \\\text{Switching to polar coordinates: } x = r \cos \theta, \, y = r \sin \theta\\I^2 = \int_0^{2\pi} \int_0^\infty e^{-r^2} \cdot r \, dr \, d\theta = \left( \int_0^{2\pi} d\theta \right) \left( \int_0^\infty r e^{-r^2} \, dr \right)\\\ \\= 2\pi \cdot \left( \frac{1}{2} \right) = \pi \Rightarrow I = \sqrt{\pi}\\\ \\\text{Hence, } \int_0^\infty e^{-x^2} \, dx = \frac{1}{2} \int_{-\infty}^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}$$

$$\frac{\sqrt{\pi}}{2}=\sqrt{\frac{\pi}{4}}$$

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